# Integer equations

I have $$2$$ following problems. Find integer roots of

\begin{align} &1)~\frac{x+y}{x^2-xy+y^2}=\frac3z \\ &2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0 \end{align}

I have no idea to solve them. I try to guess roots of the second, they are $$\left( -2, 1\right), \left( 0, -1\right), \left( 0, 3\right), \left( 2, 1\right)$$. Please help me. Thank you very much.

• These problems seem very interesting. Where did you get them from? – Toby Mak Dec 8 '18 at 9:39
• @TobyMak These are my homework. – Cglkttca Dec 8 '18 at 9:42
• What class are you taking? – Toby Mak Dec 8 '18 at 9:42
• @TobyMak I'm in grade 9. – Cglkttca Dec 8 '18 at 9:43
• What lesson is this that you were given that exercise ? – Rebellos Dec 8 '18 at 10:03

Question 1:

\begin{align} & \frac{x+y}{x^2-xy+y^2} = \frac 3z \\ \iff & \frac{3(x^2-xy+y^2)}{x+y} = z\\ \iff & 3(x+y) - \frac{9xy}{x+y} = z \\ \iff & \frac{9xy}{x+y} = 3x+3y-z \\ \implies & \frac{9}{\frac 1x + \frac 1y} = 3x+3y-z \end{align}

From the LHS we see that $$\frac 1x + \frac 1y$$ must be equal to $$\pm 1, \pm 3, \pm 9$$ since the RHS is an integer. But of course, since $$x,y$$ are integers, $$\frac 1x + \frac 1y \in [-2,2]$$ and it follows that $$\frac 1x + \frac 1y = \pm 1$$.

Moreover, the only way this can happen is if $$x = y = \pm 2$$.

Hence, the only solutions are $$(x,y,z) = (2,2,3)$$ and $$(x,y,z) = (-2,-2,-3)$$.

EDIT:

As pointed out in the comments, I have made the mistake of dividing by $$0$$, so I have changed the last $$\iff$$ to an $$\implies$$.

If $$x=0$$, then $$z=3y$$ so we get the solutions $$(x,y,z) = (0,t,3t)$$

Similarly when $$y=0$$ we get the solutions $$(x,y,z) = (t,0,3t)$$ for any $$t \in \Bbb Z$$ with $$t \neq 0$$.

• I got it. Thank you very much. – Cglkttca Dec 8 '18 at 10:20
• It's a nice answer, however, you forgot the cases when x and y are 0. – Ankit Kumar Dec 8 '18 at 10:29
• Yes, in this case, we get more solutions – Cglkttca Dec 8 '18 at 10:34